Algebra Factorisation Prof David Marshall School of Computer Science Informatics brPage 3br Factorisation The ac Method Cubic Equations Quartic Equations Factorisation Factorisation is a way of simplifying algebraic expressions As we have seen thi ID: 87006
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FactorisationTheacMethod CubicEquationsQuarticEquations Factorisation Factorisationisawayofsimplifyingalgebraicexpressions. Aswehaveseen,thiscanbee ectivewhenndingtherootsofanequationf(x)=0. Therearemanyotheruseswherefactorisationcansimplifythemathse.g.algebraicfractionsSimplefactorisationexamples: 25ab2�15a2b 40ab2�24a2b=5ab(5b�3a) 8ab(5b�3a)=5 8 x2�2x+1=(x�1)2 x2�3x+2=(x�1)(x�2)2/26 FactorisationTheacMethod CubicEquationsQuarticEquations MethodsandToolsforFactorisation Therearemanymethodsandtoolsyoucanusetofactoriseexpressions Commonfactors Commonfactorsbygrouping Theacmethod Knowyouralgebra!-PracticemakesPerfect!3/26 FactorisationTheacMethod CubicEquationsQuarticEquations Commonfactors Thesimplestformoffactorisation: Extractthehighestcommonfactors(HCF)fromanexpression Thesecanbevariablesand/orintegers(coecients).Example:24x2y2�18xy3 TheHCFofthecoecientsis6 TheHCFofxisx TheHCFofyisy2So24x2y2�18xy3=6xy2(4x�3y)4/26 FactorisationTheacMethod CubicEquationsQuarticEquations Exercise:Commonfactors Factorise: 8x4y3+6x3y2= 15a3b�9a2b2=5/26 FactorisationTheacMethod CubicEquationsQuarticEquations Commonfactorsbygrouping Multipletermedexpressionscansometimesbefactorisedintobinomialexpressionsbyextractingcommonfactorsfromeach,e.g.:2ac+6bc+ad+3bd=(2ac+6bc)+(ad+3bd)=2c(a+3b)+d(a+3b)=(a+3b)(2c+d)Thetrickhereistospotthefactors!6/26 FactorisationTheacMethod CubicEquationsQuarticEquations Exercise:Commonfactorsbygrouping Factorise: x3+4x2y�xy�4y2=7/26 FactorisationTheacMethod CubicEquationsQuarticEquations KnowyourAlgebra! Recap:GeneralRulesofAlgebraicMultiplication: c(a+b)=ac+bc (a+b)(c+d)=ac+ad+bc+bd (a+b)2=(a+b)(a+b)=aa+ba+ab+bb=a2+2ab+b2 (a�b)(a+b)=aa+ab�ba�bb=a2+ab�ab�b2=a2�b28/26 FactorisationTheacMethod CubicEquationsQuarticEquations Examples:Expanding/SimplifyingExpressions Wearequiteusedtoexpandingalgebraicexpressions: 3(6a+3b�c)�5(2a�b+3c)=18a+9b�3c�10a+5b�15c=8a+14b�18c 2(3m�n)+4(m+2n)�3(2m+3n)=6m�2n+4m+8n�6m�9n=4m�3n (x+1)(x+6)=x2+x+6x+6=x2+7x+6 (x+y)(m+n)=mx+nx+my+ny (x+4)2=x2+8x+16 (x�3)(x+3)=x2�99/26 FactorisationTheacMethod CubicEquationsQuarticEquations Examples:Factorisation Factorisationsofthefollowingexpressionsare: 8x2�12x=4x(2x�3) 5x2+15x3=5x2(1+3x) x2�2x�15=(x+3)(x�5) x2+9x+20=(x+5)(x+4) xz+2yz�2y�x=xz�x+2yz�2y=x(z�1)+2y(z�1)=(z�1)(x+2y) 2x2�3x�2=(2x+1)(x�2)10/26 FactorisationTheacMethod CubicEquationsQuarticEquations TheacMethod Guidanceforfactorisingquadraticsofthetypeax2+bx+cwherea6=0 Obtainjacji.e.thenumericalvalueoftheproductacignoringthesignoftheproduct. Writedownallthepossiblepairsoffactorsofjacj Ifcispositive,weselectthetwofactorsofjacjwhosesumisequaltojbj:bothofthesefactorshavethesamesignasb. Ifcisnegative,weselectthetwofactorsofjacjwhichdi erbythevalueofjbj;thenumericallylargerofthesetwofactorshasthesamesignasthatofbandtheotherhastheoppositesign. Ineachcase,denotethetwofactorsobtainedasf1andf2 Thenax2+bx+cisnowwrittenax2+f1x+f2x+candthisisfactorisedbyndingcommonfactors.11/26 FactorisationTheacMethod CubicEquationsQuarticEquations Example:Factorisingquadraticsax2+bx+c,a6=0 Factorise:6x2+11x+3Sowehave a=6,b=11,c=3 Thereforejacj=18 Factorsof18:(1;18);(2;9);(3;6) cis+vetherefore: f1andf2shouldadduptojbj=11 Thereforetherequiredfactorsare(2;9)Wethereforewrite:6x2+11x+3=6x2+2x+9x+3=(6x2+9x)+(2x+3)=3x(2x+3)+1(2x+3)=(3x+1)(2x+3)12/26 FactorisationTheacMethod CubicEquationsQuarticEquations Exercise:Factorisingquadraticsax2+bx+c,a6=1or0 Factorise: 3x2�14x+8= 8x2+18x�5=13/26 FactorisationTheacMethod CubicEquationsQuarticEquations FactorsasAlgebraicFractions Trivially:AnyfactorswhichappearinBOTHthenumeratoranddenominatorarecalledcommonfactorsandcansimplybecancelledSimpleexample:18x2 6x=3xThisisausefulpropertytosimilarlyapplytoalgebraicfractions: Afractionisexpressedinitssimplestformbyfactorisingthenumeratoranddenominatorandcancellinganycommonfactors.Example:Simplifyx2�1 x2+3x+2=(x�1)(x+1) (x+2)(x+1)=x�1 x+214/26 FactorisationTheacMethod CubicEquationsQuarticEquations CubicEquations Iff(x)isacubicfunctionax3+bx2+cx+dthen theequationf(x)=0canhaveuptothreerealroots.Note: Thenumberofrealrootswilldependuponthevaluesofa,b,candd. Factorsofthecubicclearlygiveusitsroots.e.g.s:x3�6x2+11x�6=0canbewrittenas(x�1)(x�2)(x�3)=0. Consequently,thevaluesofxwhichsatisfythisequationarex=1,x=2andx=315/26 FactorisationTheacMethod CubicEquationsQuarticEquations FactorTheoremandRemainderTheorem(1) FactortheoremDenition:Ifforagivenpolynomialfunctionf(x),f(a)=0thenx�aisafactorofthepolynomialf(x).Example:Factorise2x3+x2�13x+6 Ifa=1thenf(1)=2+1�13+66=0|(x�1)isnotafactor Ifa=�1thenf(�1)=�2+1+13+66=0|(x+1)isnotafactor Ifa=2thenf(2)=16+4�26+6=0|(x�2)isafactor16/26 13x+62x x2!10 FactorisationTheacMethod CubicEquationsQuarticEquations FactorTheoremandRemainderTheorem(2) Tondotherfactorsof2x3+x2�13x+6wecannowfactoroutx�2.Weneedtodoarithmetic(long)division: 17/26 FactorisationTheacMethod CubicEquationsQuarticEquations FactorTheoremandRemainderTheorem(3) Itfollowsthat2x2+5x�3isalsoafactorof2x3+x2�13x+6.Now2x2+5x�3=(2x�1)(x+3)Therefore(2x�1)(x+3)arefactorstooSoallthefactorsof2x3+x2�13x+6are:(x�2)(2x�1)(x+3)18/26 FactorisationTheacMethod CubicEquationsQuarticEquations Exercise:FactorTheorem Usethefactortheoremtofactorisethefollowingpolynomialx3+3x2�x�3:Whataretherootsoftheequation:x3+3x2�x�3=0?19/26 FactorisationTheacMethod CubicEquationsQuarticEquations FactorTheoremandRemainderTheorem(4) Remaindertheorem:Whenf(x)isdividedby(x�a)remainderisf(a).Example:Findtheremainderwhenx3+6x2+7x�4isdividedbyx+3 Usingtheremaindertheorema=�3 Sof(a)=f(x=a=3),(settingx=�3)x3+6x2+7x�4=(�3)3+6(�3)2+7(�3)�4=�27+54�21�4=220/26 FactorisationTheacMethod CubicEquationsQuarticEquations Exercise1:RemainderTheorem Whenthepolynomialf(x)=x3+8x2+kx+10isdividedbyx�2,thereisaremainderof84.Showthatk=17.21/26 FactorisationTheacMethod CubicEquationsQuarticEquations Exercise2:RemainderTheorem Giventhatx+3isafactorof2x3+9x2+ax�6,ndthevalueofa.Usingthisvalueofa,solvetheequation2x3+9x2+ax�6=0.22/26 FactorisationTheacMethod CubicEquationsQuarticEquations Sumordi erenceoftwocubes Anotherformulaforfactoringisthesumordi erenceoftwocubes.Thesumcanberepresentedbya3+b3=(a+b)(a2�ab+b2)andthedi erencebya3�b3=(a�b)(a2+ab+b2)23/26 FactorisationTheacMethod CubicEquationsQuarticEquations QuarticEquations Recap:APolynomialofOrder4:f(x)=ax4+bx3+cx2+dx+eInprincipal,thesolutiontoaQuarticusesthesametoolsasforacubicandquadratic-it'sjustalittlemorelongwindedormorecomplex. Thesameidea,asjustintroducedwithcubics,ofusingtheFactorTheoremapplies24/26 """""#a24!2b3+213(b2!3ac+12d) %!4( 254) 2+27a2d!72bd+%!4(b 2d!72bd+%!4( 254) 9abc+27c2 +12d)3+(2b3!9abc+27c2+27 !72bd+% 254) !72bd+%!4(b 2&13!'( %!4( 254) 9abc+27c2 9abc+27c2 254&1 %!4( !72bd+%!4(b 2&13!'( !2b3!9abc+27c2 "!4(b2!3ac+12d)3+(2b3!9abc+27c2 9abc+27c2 254&1 !72bd+% 254) !72bd+%!4(b 2&13!'( %!4( 254) 213(b2!3ac+12d) 9abc+27c2 9abc+27c2 254&1 """""#a24!2b3+213(b2!3ac+12d) %!4( 254) 2+27a2d!72bd+%!4(b 2d!72bd+%!4( 254) 9abc+27c2 +12d)3+(2b3!9abc+27c2+27 !72bd+% 254) !72bd+%!4(b 2&13!'( %!4( 254) 9abc+27c2 9abc+27c2 254&1 %!4( !72bd+%!4(b 2&13!'( !2b3!9abc+27c2 "!4(b2!3ac+12d)3+(2b3!9abc+27c2 9abc+27c2 254&1 !72bd+% 254) !72bd+%!4(b 2&13!'( %!4( 254) 213(b2!3ac+12d) 9abc+27c2 9abc+27c2 254&1 FactorisationTheacMethod CubicEquationsQuarticEquations ClosedformQuarticSolution Aclosedformsolutionforaquarticactuallydoesexist: Firstpartenlarged: However,inpractice,thisistoounwieldytobeusedforsolvingquarticequations(Sourcehttp://planetmath.org/encyclopedia/QuarticFormula.html )25/26 FactorisationTheacMethod CubicEquationsQuarticEquations MatlabtotheRescue WecanuseMATLABtorelievethestressoffactorisation: ��symsx;��f=4x^4�3x^3�2x^2+x;��factor(f)ans=x(x�1)(4x^2+x�1)��double(solve(f))ans=01.00000.3904�0.6404 26/26 CM2202:ScienticComputingandMultimediaApplicationsGeneralMaths:2.Algebra-FactorisationProf.DavidMarshallSchoolofComputerScience&Informatics